Solution of the equation $\left| {\,\begin{array}{*{20}{c}}1&1&x\\{p + 1}&{p + 1}&{p + x}\\3&{x + 1}&{x + 2}\end{array}\,} \right| = 0$ are
$x = 1,\,2$
$x = 2,\,3$
$x = 1,\,p,\,2$
$x = 1,\,2,\, - p$
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right| = $
If the system of linear equations $2 x+3 y-z=-2$ ; $x+y+z=4$ ; $x-y+|\lambda| z=4 \lambda-4$ (where $\lambda \in R$), has no solution, then
If $\left| {\,\begin{array}{*{20}{c}}a&b&c\\m&n&p\\x&y&z\end{array}\,} \right| = k$, then $\left| {\,\begin{array}{*{20}{c}}{6a}&{2b}&{2c}\\{3m}&n&p\\{3x}&y&z\end{array}\,} \right| = $
If $a,b,c$ be positive and not all equal, then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|$ is
If $\omega $ is an imaginary root of unity, then the value of $\left| {\,\begin{array}{*{20}{c}}a&{b{\omega ^2}}&{a\omega }\\{b\omega }&c&{b{\omega ^2}}\\{c{\omega ^2}}&{a\omega }&c\end{array}\,} \right|$ is